ode question from a textbook - probably an easy answer

Hi,

I do not understand the following derivation. Could somebody please enlighten me:

dx/dt = b - ax

to solve, define a new variable y(t) = x(t) - b/a

since dy/dt = dx/dt, we can derive the equation dy/dt = - ay [*]

since y(t) = y(0)exp(-at) and substituting back to express the solution in terms of x(t)

x(t) = b/a + (x(0) - b/a) exp(-at) [*]

>>>I don't know how they got [*] and then [**].

Re: ode question from a textbook - probably an easy answer

Quote:

Originally Posted by

**selecttext** Hi,

I do not understand the following derivation. Could somebody please enlighten me:

dx/dt = b - ax

to solve, define a new variable y(t) = x(t) - b/a

since dy/dt = dx/dt, we can derive the equation dy/dt = - ay [*]

since y(t) = y(0)exp(-at) and substituting back to express the solution in terms of x(t)

x(t) = b/a + (x(0) - b/a) exp(-at) [*]

>>>I don't know how they got [*] and then [**].

Using

dx/dt = b - ax

replace dx/dt with dy/dt and replace x with (y + b/a)

dy/dt = b - a(x + b/a) = -ay

using

y(t) = x(t) - b/a

solve for x(t) and substitute for y(t) from THEIR previous line.

I think the (t) symbol may be confusing...

:)

Re: ode question from a textbook - probably an easy answer

sweet thanks! i am revisiting an old textbook and i once knew this solution. what a revelation

Re: ode question from a textbook - probably an easy answer

Quote:

Originally Posted by

**selecttext** Hi,

I do not understand the following derivation. Could somebody please enlighten me:

dx/dt = b - ax

to solve, define a new variable y(t) = x(t) - b/a

since dy/dt = dx/dt, we can derive the equation dy/dt = - ay [*]

since y(t) = y(0)exp(-at) and substituting back to express the solution in terms of x(t)

x(t) = b/a + (x(0) - b/a) exp(-at) [*]

>>>I don't know how they got [*] and then [**].

I don't know why they're bothering to bring in new variables when it's already first order linear and can be solved easily using an integrating factor.

Now multiplying both sides by the integrating factor we find